【Python+Stata】豪斯曼检验：固定效应or随机效应？
文章目录
- 一、Python实现
-
- （一）获取面板数据
- （二）固定效应模型
- （三）随机效应模型
- （四）豪斯曼检验
- 二、Stata实现
-
- （一）获取面板数据
- （二）固定效应模型
- （三）随机效应模型
- （四）豪斯曼检验
本文研究陈强《计量经济学及Stata应用》12.13案例的python及stata实现。lin_1992dta数据集链接：链接：https://pan.baidu.com/s/10WAjBbZgL4NzZV69JbgbZw 提取码：d0dx
一、Python实现
（一）获取面板数据
import numpy as np
import linearmodels as plm
import scipy.stats as stats
import pandas as pd
lin =pd.read_stata(r'C:\Users\mi\Documents\stata\lin_1992.dta')
lin = lin.set_index(['province', 'year'], drop=False)
（二）固定效应模型
reg_fe = plm.PanelOLS.from_formula(formula='ltvfo ~1 +ltlan + ltwlab + ltpow + ltfer + hrs + mci+ ngca + EntityEffects',data=lin)
results_fe = reg_fe.fit()
（三）随机效应模型
reg_re = plm.RandomEffects.from_formula(
formula='ltvfo ~1 +ltlan + ltwlab + ltpow + ltfer + hrs + mci+ ngca', data=lin)
results_re = reg_re.fit()
（四）豪斯曼检验
b_fe = results_fe.params
b_re = results_re.params
b_diff = b_fe - b_re
v_fe = results_fe.cov
v_re = results_re.cov
v_diff = v_fe - v_re
df = len(b_fe)
table = pd.DataFrame({'FE':b_fe,'RE':b_re,'Difference':b_diff,'sqrt(diag(v_fe-v_re))':np.sqrt(np.diag(v_diff))})
chi2 = np.dot(b_diff.T, np.linalg.inv(v_diff).dot(b_diff))
pval = 1 - stats.chi2.cdf(chi2, df)
print(table)
print()
print(f'chi-Squared: {chi2:.2f}')
print(f'degrees of freedom: {df}')
print(f'p-Value:{pval:.5f}')
结果为：
FE RE Difference sqrt(diag(v_fe-v_re))
Intercept 2.310114 2.387878 -0.077764 0.178212
hrs 0.207582 0.218610 -0.011028 NaN
ltfer 0.176277 0.188274 -0.011997 0.004112
ltlan 0.639966 0.565591 0.074374 0.043153
ltpow 0.077160 0.060477 0.016683 0.001574
ltwlab 0.123993 0.144184 -0.020192 0.009747
mci 0.258036 0.470237 -0.212201 0.052192
ngca 0.772279 0.674517 0.097762 0.046883
chi-Squared: 48.68
degrees of freedom: 8
p-Value:0.00000
二、Stata实现
（一）获取面板数据
use C:\Users\mi\Documents\stata\lin_1992.dta,clear
xtset province year
结果为：
Panel variable: province (strongly balanced)
Time variable: year, 70 to 87
Delta: 1 unit
（二）固定效应模型
xtreg ltvfo ltlan ltwlab ltpow ltfer hrs mci ngca,fe
estimates store FE
（三）随机效应模型
xtreg ltvfo ltlan ltwlab ltpow ltfer hrs mci ngca,re
estimates store RE
（四）豪斯曼检验
由于传统的豪斯曼检验假设球形扰动项，故在进行固定效应与随机效应的估计时，均不使用异方差或聚类稳健的标准误。
hausman FE RE,constant
其中，选择项“constant”表示在比较系数估计值时包括常数项（默认不包括常数项）
结果为：
---- Coefficients ----
| (b) (B) (b-B) sqrt(diag(V_b-V_B))
| FE RE Difference Std. err.
-------------+----------------------------------------------------------------
ltlan | .6399658 .5655915 .0743743 .0431529
ltwlab | .1239927 .1441844 -.0201917 .009747
ltpow | .0771604 .060477 .0166834 .001574
ltfer | .1762775 .1882741 -.0119966 .0041117
hrs | .2075817 .2186096 -.0110279 .
mci | .2580359 .4702368 -.2122009 .052192
ngca | .7722795 .6745175 .097762 .0468832
_cons | 2.310114 2.387878 -.0777638 .178212
------------------------------------------------------------------------------
b = Consistent under H0 and Ha; obtained from xtreg.
B = Inconsistent under Ha, efficient under H0; obtained from xtreg.
Test of H0: Difference in coefficients not systematic
chi2(8) = (b-B)'[(V_b-V_B)^(-1)](b-B)
= 48.68
Prob > chi2 = 0.0000
(V_b-V_B is not positive definite)
由于p指为0.00000，故强烈拒绝原假设，认为应该使用固定效应模型，而非随机效应模型。
但是很多时候计算出的 s q r t ( d i a g ( V b − V B ) ) sqrt(diag(V_b-V_B)) sqrt(diag(Vb−VB))可能为负，使用sigmamore选项，表示统一使用随机效应估计量的方差估计，可以大大减少出现负值的可能性。
使用sigmamore选项，我不知道怎么在python中实现，可能还是不懂sigmamore的实现原理。先mark一下，以后再更新。
hausman FE RE,constant sigmamore
结果为：
---- Coefficients ----
| (b) (B) (b-B) sqrt(diag(V_b-V_B))
| FE RE Difference Std. err.
-------------+----------------------------------------------------------------
ltlan | .6399658 .5655915 .0743743 .0476709
ltwlab | .1239927 .1441844 -.0201917 .0125808
ltpow | .0771604 .060477 .0166834 .0081232
ltfer | .1762775 .1882741 -.0119966 .0078425
hrs | .2075817 .2186096 -.0110279 .0052769
mci | .2580359 .4702368 -.2122009 .0583709
ngca | .7722795 .6745175 .097762 .0828671
_cons | 2.310114 2.387878 -.0777638 .2078242
------------------------------------------------------------------------------
b = Consistent under H0 and Ha; obtained from xtreg.
B = Inconsistent under Ha, efficient under H0; obtained from xtreg.
Test of H0: Difference in coefficients not systematic
chi2(8) = (b-B)'[(V_b-V_B)^(-1)](b-B)
= 48.49
Prob > chi2 = 0.0000
(V_b-V_B is not positive definite)